" data-orsrc="https://uploads-pracbee-ramanuj.s3.ap-south-1.amazonaws.com/pyq-question-images/batch-001/09730c56dbe8.webp" loading="lazy" style="max-width:100%;height:auto;display:block;margin:0 auto;max-height:40vh;vertical-align:baseline" alt="JEE Main 2024 (Online) 6th April Evening Shift Mathematics - Properties of Triangle Question 3 English Explanation">

$$\begin{aligned} & \cos A=\frac{2}{3}=\frac{\alpha^2+8^2-7^2}{2 \cdot \alpha \cdot 8} \\ & \Rightarrow\left(\alpha^2+15\right) 3=32 \alpha \\ & 3 \alpha^2-32 \alpha+45=0 \\ & \Rightarrow \alpha=\frac{5}{3}, 9 \\ & \because \alpha \in N \Rightarrow \alpha=9 \\ & \cos C=\frac{7^2+8^2-9^2}{2 \cdot 7 \cdot 8}=\frac{2}{7} \end{aligned}$$

$$ \begin{aligned} \cos 3 C & =4 \cos ^3 C-3 \cos C \\ = & \frac{4 \times 8}{7^3}-\frac{6}{7} \\ 49 \cos 3 C & =\frac{32}{7}-42 \\ \Rightarrow \quad & 49 \cos 3 C+42=\frac{32}{7} \\ \Rightarrow \quad & m+n =39 \end{aligned} $$

19:T451,

JEE Main 2024 (Online) 6th April Evening Shift Mathematics - Properties of Triangle Question 3 Hindi Explanation

$$ \begin{aligned} \cos 3 C & =4 \cos ^3 C-3 \cos C \\ = & \frac{4 \times 8}{7^3}-\frac{6}{7} \\ 49 \cos 3 C & =\frac{32}{7}-42 \\ \Rightarrow \quad & 49 \cos 3 C+42=\frac{32}{7} \\

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